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3 years ago

The formula for calculating gravitational potential energy is PE=______?

2 answers:

3 0

La Energía potencial es la energía que tiene un cuerpo situado a una determinada altura sobre el suelo. Ej.: El agua embalsada, que se manifiesta al caer y mover la hélice de una turbina. La energía potencial, Ep, se mide en julios (J), la masa, m se mide en kilogramos (kg), la aceleración de la gravedad, g, en metros/segundo-cuadrado (m/s2) y la altura, h, en )m(

Aneli [31]3 years ago

3 0

The formula for potential energy depends on the force acting on the two objects. For the gravitational force the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m / s2 at the surface of the earth) and h is the height in meters.

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Use the Pythagorean Theorem to find the velocity of an airplane traveling north at 100 kilometers per hour. It is also flying th

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3 years ago

g determine what frequency is required of a source powering a 100 uf capacitor a 500 ohm resistor and a s50 mH inductor in serie

polet [3.4K]

Answer: $71.16\ Hz$

Explanation:

Given

Capacitance $C=100\ \mu F$

Resistance $R=500\ \Omega$

Inductance $L=50\ mH$

In LCR circuit, current is maximum at resonance frequency i.e.

$X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}$

Insert the values

$\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}$

Also, frequency is given by

$\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}$

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8 0

3 years ago

Manuel releases a pendulum bob. The drawing shows its path. At which point does the pendulum bob have the most gravitational pot

kati45 [8]

We have no clue until we see the drawing. After all, the drawing shows its path.

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3 years ago

In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0580 s, during which

Fantom [35]

Answer:

V₀ₓ = 10.94 m/s

V₀y = 18.87 m/s

Explanation:

To find the launch velocity, we use 1st equation of motion.

Vf = Vi + at

where,

Vf = Final Velocity of Ball = Launch Speed = V₀ = ?

Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)

a = acceleration = 376 m/s²

t = time = 0.058 s

Therefore,

V₀ = 0 m/s + (376 m/s²)(0.058 s)

V₀ = 21.81 m/s

Now, for x-component:

V₀ₓ = V₀ Cos θ

where,

V₀ₓ = x-component of launch velocity = ?

θ = Angle with horizontal = 59.9⁰

V₀ₓ = (21.81 m/s)(Cos 59.9°)

V₀ₓ = 10.94 m/s

for y-component:

V₀ₓ = V₀ Sin θ

where,

V₀y = y-component of launch velocity = ?

θ = Angle with horizontal = 59.9⁰

V₀y = (21.81 m/s)(Sin 59.9°)

V₀y = 18.87 m/s

8 0

4 years ago